# Lebesgue measure as $\sup$ of measures of contained compact sets

I know, from Kolmogorov-Fomin's Элементы теории функций и функционального анализа, the definition of external measure of a bounded set $A\subset \mathbb{R}^n$ as $$\mu^{\ast}(A):=\inf_{A\subset \bigcup_k P_k}\sum_k m(P_k)$$where the infimum is extended to all the possible covers of $A$ by finite or countable families of $n$-paralleliped $P_k=\prod_{i=1}^n I_i$, where $I_i=[a_{i,k},b_{i,k}]$ or $I_i=(a_{i,k},b_{i,k})$ or $I_i=(a_{i,k},b_{i,k}]$ or $I_i=[a_{i,k},b_{i,k})$, with $a_{i,k}\le b_{i,k}$, whose measure is defined as $$m(P_k):=\prod_{i=1}^n(b_{i,k}-a_{i,k}).$$ A set is said to be elementary when it is the union of a finite number of such $n$-parallelipeds and a set $A$ is said to be measurable if, for any $\varepsilon>0$, there is an elementary set $B$ such that $$\mu^{\ast}(A\triangle B)<\varepsilon.$$The function $\mu^{\ast}$ defined on measurable sets only is called Lebesgue measure and the notation $\mu$ is used for it.

Let us come to my question. I read that if $A$ is Lebesgue measurable then $$\mu(A)=\sup_{K\subset A} \mu(K)$$ where the $K$ are compact measurable sets. My text, Kolomogorv-Fomin's does not examine the theory of Lebesgue measure in $\mathbb{R}^n$, $n>1$, in depth, therefore I have no tool to prove such a property to myself, and I would be very grateful to anybody explaining a proof to me.

• The definition you give of measurability is wrong. What do K&M really give for the definition? What you say is fine for sets with $\mu^*$ finite, but for example the set $\{(x,y)\in\Bbb R^2:|y-x|<1\}$ is certainly Lebesgue measurable, but it does not satisfy the condition you cite. – David C. Ullrich May 29 '16 at 17:26
• @DavidC.Ullrich Specified: the focus of Kolmogorov and Fomin's book is more on finite measure and it uses $A$ bounded to define $\mu$. – Self-teaching worker May 29 '16 at 19:13

Assuming we've shown that $\mu$ is a measure, and that we're only concerned with bounded sets:

Say $B$ is a bounded closed set and $E\subset B$ is measurable. First, note that $$\mu(E)=\inf\{\mu(V):E\subset V\subset B, V\text{ relatively open}\}.$$

(Here "relatively open" means open relative to $B$; that is, $V=B\cap W$ where $W$ is an open subset of $\Bbb R^d$.)

The inequality $\le$ is clear. Let $\epsilon>0$, and choose rectangles $I_j$ with $E\subset \bigcup I_j$ and $\sum\mu(I_j)<\mu(E)+\epsilon$. Let $O_j$ be an open rectangle with $I_j\subset O_j$ and $$\mu(O_j)<\mu(I_j)+\epsilon2^{-j}.$$Let $V=B\cap\bigcup O_j$. Then $V$ is relatively open, $E\subset V$, and $$\mu(V)\le\sum\mu(O_j)\le \epsilon+\sum\mu(I_j)<\mu(E)+2\epsilon.$$ So $\mu(E)\ge\inf_V\mu(V)$, hence $\mu(E)=\inf_V\mu(V)$.

That's known as outer regularity. Inner regularity follows by taking complements: Given $E\subset B$, let $F=B\setminus E$. There exists a relatively open set $V\subset B$ with $F\subset V$ and $\mu(V)<\mu(F)+\epsilon$. Let $K=B\setminus V$. Then $K$ is compact, $F\subset V$ implies $K\subset E$, and $$\mu(K)=\mu(B)-\mu(V)>\mu(B)-\mu(F)-\epsilon=\mu(E)-\epsilon.$$

• Thank you so much: very clear and complete answer! I think the only needed assumption of $E$ is measurability (not necessarily openness), am I right? Moreover, I think it is intended that bounded open (equivalently closed) sets are Lebesgue measurable: am I right (in that case I think it may be the interesting topic of another question)? – Self-teaching worker May 30 '16 at 15:51
• @Self-teachingworker I left out the main hypothesis, sorry. Yes, $E$ is measurable. Yes, open sets are measurable. For bounded open sets that's immediate from the definition plus that lemma elsewhere that a bounded open set is the union of almost disjoint rectangles (apply the definition of measurability to the union of the first $N$ of those rectangles). It's possible you'd really be happier simply finding another book - all this is very standard (to tell you the truth I'm surprised the current question hasn't got any negative votes on it's-in-the-book grounds). Royden might work... – David C. Ullrich May 30 '16 at 16:01
• Very clear and exhaustive: I heartily thank you. As to Kolmogorov-Fomin's, it's often considered a classic, but yours isn't the first suggestion to use another book to study measure theory and functional analysis I receive. I've finished it and I'd like to deepen my knowledge (I'm a self-teaching student, as said in my profile) of such topics on another text, so I'm very grateful to you for your suggestion! – Self-teaching worker May 31 '16 at 18:05